Worksheet on Mensuration

Solved Questions on Mensuration

1. If the length and the breadth of a rectangular paper are 37 cm and 21 cm, what is the area of the largest circle which can be cut out of this paper?

a) 346.5 cm²
b) 537.82 cm²
c) 693 cm²
d) 1075.64 cm²

Answer: a) 346.5 cm²

Explanation: Length (l) of a rectangular paper = 37 cm
Breadth (b) of a rectangular paper = 21 cm

Largest circle which can be cut out of this paper with the largest diameter (d) is equal to the breadth (b) of a rectangular paper.
d = b
r = d/2 = b/2
= 21/2 cm

Area of a largest circle = 22/7 × (21/2)² [Area of a circle = πr²,   Where π = 22/7 ]
= 22/7 × (21/2) × (21/2)
= 346.5 cm²

2. In a square field, a quarter of a circle with a radius of 7 centimetres is designated as a water pool starting from each of its corners. What is the area of the remaining part of the grassy area within the field if each side of the field is 35 metres?

a) 1051 m²
b) 1061 m²
c) 1071 m²
d) 1081 m²

Answer: c) 1071 m²

Explanation: Area of a square field = a² = 35² = 1225 m²

Four quadrants form a circle.

Area of four quadrants = Area of a circle [Area of a circle = πr²,   Where π = 22/7]
= 22/7 × (7)²
= 22/7 × 7 × 7
= 154 m²
Area of the remaining part of the grassy area within the field
= Area of a square field − Area of four quadrants
= (1225 − 154) m²
= 1071 m²

3. The circumference of a circular park is 264 metres and the width of the footpath around the park is 14 metres. what is the cost of fencing the outer circumference at the rate of $5.82 per metre?

a) $2028.24
b) $2028.64
c) $2048.24
d) $2048.64     

Answer: d) $2048.64

Explanation: The figure is:

Inner circumference of a circular park = 264 m
⇒ 2πr = 264 m
⇒ 2 × 22/7 × r = 264 m [π = 22/7]  
⇒ r = 264 × ½ × 7/22
⇒ r = 264 × ½ × 7/22
⇒ r = 42 m

Inner radius = r = 42 m

Outer radius = R =  r + 14

= 42 + 14 
= 56 m 

Outer circumference of a footpath around a circular park = 2πR
= 2 × 22/7 × 56 
= 352 m

Cost of fencing the outer circumference of 1 m = $5.82

Cost of fencing the outer circumference of 352 m = $5.82 × 352
                                                                         = $2048.64

4. What is the total surface area of a cuboid if the side of a small cube is 3 cm?

a) 428 cm
b) 448 cm
c) 468 cm
d) 488 cm

Answer: c) 468 cm

Explanation: Side of a small cube = 3 cm
Length (l) of a cuboid = 4 × Side of a small cube = 4 × 3 = 12 cm
Breadth (b) of a cuboid = 2 × Side of a small cube = 2 × 3 = 6 cm
Height (h) of a cuboid = 3 × Side of a small cube = 3 × 3 = 9 cm  
Total surface area of a cuboid = 2(lb + bh + hl)
= 2(12 × 6 + 6 × 9 + 9 × 12)
= 2(72 + 54 + 108)
= 2 × 234
= 468 cm

5. The radius and height of the pig iron cylindrical roll are 49 centimetres and 484 centimetres, respectively. The pig iron is melted in the furnace and a huge cube is formed. What is the lateral surface of a cube formed?

a) 7.5 m²
b) 9.5 m²
c) 11.5 m²
d) 13.5 m²

Answer: b) 9.5 m²

Explanation: r = 7 cm

h = 14 cm

Volume of cylinder = πr²h
= 22/7 × 49² × 484
= 22/7 × 49 × 49 × 22 × 22
= 22/7 × 49 × 49 × 22 × 22
= (22 × 22 × 22 × 7 × 7 × 7) cm³

Volume of an iron cube formed = Volume of cylinder
⇒ a³ = 22 × 22 × 22 × 7 × 7 × 7
⇒ a = 22 × 7
⇒ a = 154 cm

Lateral surface area = 4 × a²
= 4 × (154)²
= 4 × 154 × 154
= 94864 cm²

Lateral surface area in m² = 94864/(100 × 100) m²
= 9.4864 m²
= 9.5 m² [Round off]